A cylinder contains 1 kg of a certain fluid at an initial pressure of 20 bar. The fluid is allowed to expand reversibly behind a piston according to a law pV2 = constant until the volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston regains its original position ; heat is then supplied reversibly with the piston firmly locked in position until the pressure rises to the original value of 20 bar. Calculate the net work done by the fluid, for an initial volume of 0.05 m3.
Mass of fluid, m = 1 kg
= `"p"_1` = 20 bar = 20 × `10_5` N/`"m"^2`
= `"V"_1` = 0.05 `"m"^3`
Considering the process 1-2
= `"P"_{1} "V"_{1}^{2}` = `"P"_{2} "V"_{2}^{2}`
= `\therefore` `"P"_{2}` = `"P"_{1}`
= `(\frac{"V"_{1}}{"V"_{2}})` = `20 (\frac{"V"_{1}}{"2V"_{2}})^2`
= `\frac{20}{4}` = 5 bar
Work done by the fluid from 1 to 2 = Area 12 `"ML"_1` = ` f_{1}^{2}` pdV
i.e. `"W"_{1-2}` =`f_{1}^{2} \frac{c}{V^{2}}dV`
where C = `"P"_{1} "V"_{1} ^{2}` = `20\times0.05^2 "bar" "m"^6`
`"W"_{1-2}` = `10^{25}\times20\times0.0025` `[-\frac{1}{V}]^{0.1}`
`"W"_{1-2}` = `10^{25}\times20\times0.0025` `(\frac{1}{0.05}-\frac{1}{0.1}) = 5000 "Nm"`
Work done on fluid from 2 to 3
= Area 32ML3 = `"p"_2` (`"V"_2` – `"V"_3`) = `10^5` × 5 × (0.1 – 0.05) = 25000 Nm
Work done during the process 3-1 = 0,
because piston is locked in position (i.e., Volume remains constant)
Net work done by the fluid
= Enclosed area 1231 = 50000 – 25000 = 25000 Nm.
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